For the equilibrium, `A(g)hArrB(g),DeltaH` is `-40kJ//mol`, if the ratio of the activation energies of the forward `(E_(f))` and reverse `(E_(b))` reactions is 2/3, thenA. `E_(f)=80kJ//mol,E_(b)=120kJ//mol`B. `E_(f)=60kJ//mol,E_(b)=100kJ//mol`C. `E_(f)=30kJ//mol,E_(b)=70kJ//mol`D. `E_(f)=70kJ//mol,E_(b)=30kJ//mol`

For the equilibrium, `A(g)hArrB(g),DeltaH` is `-40kJ//mol`, if the rat

1.	For the equilibrium, `A(g)hArrB(g),DeltaH` is `-40kJ//mol`, if the ratio of the activation energies of the forward `(E_(f))` and reverse `(E_(b))` reactions is 2/3, thenA. `E_(f)=80kJ//mol,E_(b)=120kJ//mol`B. `E_(f)=60kJ//mol,E_(b)=100kJ//mol`C. `E_(f)=30kJ//mol,E_(b)=70kJ//mol`D. `E_(f)=70kJ//mol,E_(b)=30kJ//mol`
Answer» Correct Answer - A Given, `A(g) hArr B(g)" "DeltaH=-40kJ//mol` So, `(E_(f))/(E_(b))=(2)/(3)` We know that, `E_(f)-E_(b)=DeltaH` `E_(f)-E_(b)=-40` `E_(b)xx(2)/(3)-E_(b)=-40` Hence, `E_(b)=120kJ//mol,E_(f)=80kJ//mol`

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